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**NCERT solutions for Class 8 Maths Comparing Quantities Download as PDF**

## NCERT Solutions for Class 8 Maths Comparing Quantities

**Class –VIII Mathematics (Ex. 8.3)**

**NCERT SOLUTION**

**1. Calculate the amount and compound interest on:**

**(a) Rs.10,800 for 3 years at **** per annum compounded annually.**

**(b) Rs.18,000 for **** years at 10% per annum compounded annually.**

**(c) Rs.62,500 for **** years at 8% per annum compounded annually.**

**(d) Rs.8,000 for **** years at 9% per annum compounded half yearly. (You could the year by year calculation using S.I. formula to verify).**

**(e) Rs.10,000 for **** years at 8% per annum compounded half yearly.**

**Ans. (a) **Here, Principal (P) = Rs. 10800, Time = 3 years,

Rate of interest (R) =

Amount (A) =

= =

= =

=

= Rs. 15,377.34

Compound Interest (C.I.) = A – P

= Rs. 10800 – Rs. 15377.34 = Rs. 4,577.34

**(b)** Here, Principal (P) = Rs. 18,000, Time = years, Rate of interest (R)

= 10% p.a.

Amount (A) =

= =

= =

= Rs. 21,780

Interest for years on Rs. 21,780 at rate of 10% = = Rs. 1,089

Total amount for years

= Rs. 21,780 + Rs. 1089 = Rs. 22,869

Compound Interest (C.I.) = A – P

= Rs. 22869 – Rs. 18000 = Rs. 4,869

**(c)** Here, Principal (P) = Rs. 62500, Time = = years = 3 years (compounded half yearly)

Rate of interest (R) = 8% = 4% (compounded half yearly)

Amount (A) =

=

=

=

=

= Rs. 70,304

Compound Interest (C.I.) = A – P

= Rs. 70304 – Rs. 62500 = Rs. 7,804

**(d)** Here, Principal (P) = Rs. 8000, Time = 1 years = 2 years(compounded half yearly)

Rate of interest (R) = 9% = (compounded half yearly)

Amount (A) =

=

=

=

=

= Rs. 8,736.20

Compound Interest (C.I.) = A – P

= Rs. 8736.20 – Rs. 8000

= Rs. 736.20

**(e)** Here, Principal (P) = Rs. 10,000, Time = 1 years = 2 years (compounded half yearly)

Rate of interest (R) = 8% = 4% (compounded half yearly)

Amount (A) =

=

=

=

=

= Rs. 10,816

Compound Interest (C.I.) = A – P

= Rs. 10,816 – Rs. 10,000 = Rs. 816

NCERT Solutions for Class 8 Maths Exercise 8.3

**2. Kamala borrowed Rs.26,400 from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?**

**(Hint: Find A for 2 years with interest is compounded yearly and then find SI on the 2 ^{nd} year amount for years).**

**Ans. **Here, Principal (P) = Rs. 26,400, Time = 2 years 4 months, Rate of interest (R) = 15% p.a.

Amount for 2 years (A) =

= =

= =

= Rs. 34,914

Interest for 4 months = years at the rate of 15% =

= Rs. 1745.70

Total amount = Rs. 34,914 + Rs. 1,745.70

= Rs. 36,659.70

NCERT Solutions for Class 8 Maths Exercise 8.3

**3. Fabina borrows Rs.12,500 per annum for 3 years at simple interest and Radhaborrows the same amount for the same time period at 10% per annnum, compounded annually. Who pays more interest and by how much?**

**Ans. **Here, Principal (P) = Rs.12,500, Time = 3 years, Rate of interest (R)

= 12% p.a.

Simple Interest for Fabina =

= = Rs. 4,500

Amount for Radha, P = Rs. 12,500, R = 10% and = 3 years

Amount (A) =

= =

= =

= Rs. 16,637.50

C.I. for Radha = A – P

= Rs. 16,637.50 – Rs. 12,500 = Rs. 4,137.50

Here, Fabina pays more interest

= Rs. 4,500 – Rs. 4,137.50 = Rs. 362.50

NCERT Solutions for Class 8 Maths Exercise 8.3

**4. I borrowsRs.12,000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay?**

**Ans. **Here, Principal (P) = Rs.12,000, Time = 2 years, Rate of interest (R) = 6% p.a.

Simple Interest =

= = Rs. 1,440

Had he borrowed this sum at 6% p.a., then

Compound Interest =

=

=

=

=

= Rs. 13,483.20 – Rs. 12,000

= Rs. 1,483.20

Difference in both interests

= Rs. 1,483.20 – Rs. 1,440.00 = Rs. 43.20

NCERT Solutions for Class 8 Maths Exercise 8.3

**5. Vasudevan invested Rs.60,000 at an interest rate of 12% per ann^um compounded half yearly. What amount would he get:**

**(i) after 6 months?**

**(ii) after 1 year?**

**Ans. (i)** Here, Principal (P) = Rs. 60,000,

Time = 6 months = 1 year(compounded half yearly)

Rate of interest (R) = 12% = 6% (compounded half yearly)

Amount (A) =

=

=

=

=

= Rs. 63,600

After 6 months Vasudevan would get amount Rs. 63,600.

**(ii)** Here, Principal (P) = Rs. 60,000,

Time = 1 year = 2 year(compounded half yearly)

Rate of interest (R) = 12% = 6% (compounded half yearly)

Amount (A) =

=

=

=

=

= Rs. 67,416

After 1 year Vasudevan would get amount Rs. 67,416.

NCERT Solutions for Class 8 Maths Exercise 8.3

**6. Arif took a loan of Rs.80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after **** years if the interest is:**

**(i) compounded annually.**

**(ii) compounded half yearly.**

**Ans. (i)** Here, Principal (P) = Rs. 80,000, Time = years, Rate of interest (R) = 10%

Amount for 1 year (A) =

=

=

=

= Rs. 88,000

Interest for year =

= Rs. 4,400

Total amount = Rs. 88,000 + Rs. 4,400 = Rs. 92,400

**(ii)** Here, Principal (P) = Rs.80,000,

Time = year = 3year (compounded half yearly)

Rate of interest (R) = 10% = 5% (compounded half yearly)

Amount (A) =

=

=

=

=

= Rs. 92,610

Difference in amounts

= Rs. 92,610 – Rs. 92,400 = Rs. 210

NCERT Solutions for Class 8 Maths Exercise 8.3

**7. Maria invested Rs.8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find:**

**(i) The amount credited against her name at the end of the second year.**

**(ii) The interest for the third year.**

**Ans. (i)** Here, Principal (P) = Rs. 8000, Rate of Interest (R) = 5%, Time = 2 years

Amount (A) =

=

=

=

=

= Rs. 8,820

**(ii)** Here, Principal (P) = Rs. 8000, Rate of Interest (R) = 5%, Time = 3 years

Amount (A) =

=

=

=

=

= Rs. 9,261

Interest for 3^{rd} year = A – P

= Rs. 9,261 – Rs. 8,820 = Rs. 441

NCERT Solutions for Class 8 Maths Exercise 8.3

**8. Find the amount and the compound interest on Rs.10,000 for **** years at 10% per annum, compounded half yearly.**

**Would this interest be more than the interest he would get if it was compounded annually?**

**Ans. **Here, Principal (P) = Rs. 10000, Rate of Interest (R) = 10% = 5% (compounded half yearly)

Time = years = 3 years (compounded half yearly)

Amount (A) =

=

=

=

=

= Rs. 11,576.25

Compound Interest (C.I.) = A – P

= Rs. 11,576.25 – Rs. 10,000 = Rs. 1,576.25

If it is compounded annually, then

Here, Principal (P) = Rs. 10000, Rate of Interest (R) = 10%, Time = years

Amount (A) for 1 year =

=

=

=

=

= Rs. 11,000

Interest for year = = Rs. 550

Total amount = Rs. 11,000 + Rs. 550

= Rs. 11,550

Now, C.I. = A – P = Rs. 11,550 – Rs. 10,000

= Rs. 1,550

Yes, interest Rs. 1,576.25 is more than Rs. 1,550.

NCERT Solutions for Class 8 Maths Exercise 8.3

**9. Find the amount which Ram will get on Rs.4,096, if he gave it for 18 months at **** per annum, interest being compounded half yearly.**

**Ans. **Here, Principal (P) = Rs. 4096,

Rate of Interest (R) =

= (compounded half yearly)

Time = 18 months = years = 3 years (compounded half yearly)

Amount (A) =

=

=

=

=

= Rs. 4,913

NCERT Solutions for Class 8 Maths Exercise 8.3

**10. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum.**

**(i) Find the population in 2001.**

**(ii) What would be its population in 2005?**

**Ans.** **(i) **Here, A_{2003} = Rs. 54,000, R = 5%, = 2 years

Population would be less in 2001 than 2003 in two years.

Here population is increasing.

A_{2003} =

54000 =

54000 =

54000 =

54000 =

48,980 (approx.)

**(ii)** According to question, population is increasing. Therefore population in 2005,

A_{2005} =

=

=

=

=

= 59,535

Hence population in 2005 would be 59,535.

NCERT Solutions for Class 8 Maths Exercise 8.3

**11. In a laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.**

**Ans. **Here, Principal (P) = 5,06,000, Rate of Interest (R) = 2.5%, Time = 2 hours

After 2 hours, number of bacteria,

Amount (A) =

=

=

=

=

=

= 5,31,616.25

Hence, number of bacteria after two hours are 531616 (approx.).

NCERT Solutions for Class 8 Maths Exercise 8.3

**12. A scooter was bought at Rs.42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.**

**Ans. **Here, Principal (P) = Rs. 42,000, Rate of Interest (R) = 8%, Time = 1 years

Amount (A) =

=

=

=

=

= Rs. 38,640

Hence, the value of scooter after one year is Rs. 38,640.

## NCERT Solutions for Class 8 Maths Exercise 8.3

NCERT Solutions Class 8 Mathematics PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 8 Mathematics includes text book solutions from Class 8 Maths Book . NCERT Solutions for CBSE Class 8 Maths have total 16 chapters. 8 Maths NCERT Solutions in PDF for free Download on our website. Ncert class 8 solutions PDF and Maths ncert class 8 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.

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